Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MXN = 1e5 + 100;
bool mark[MXN];
int dp[MXN][18];
int dis[MXN];
vector<int> g[MXN];
vector<int> pd[MXN][18];
vector<int> pdd[MXN];
int n, lgn, m, q;
int lg(int a) {
  int l = 0;
  int aa = 1;
  while (aa < a) {
    aa *= 2;
    l++;
  }
  return l;
}
void merge(vector<int> v, vector<int> w, vector<int> &vw) {
  while (vw.size()) vw.pop_back();
  int p1 = 0, p2 = 0;
  int sz = v.size() + w.size();
  sz = min(sz, 10);
  for (int i = 0; i < sz && p1 < v.size() && p2 < w.size(); i++) {
    if (v[p1] > w[p2]) {
      vw.push_back(w[p2]);
      p2++;
    } else {
      vw.push_back(v[p1]);
      p1++;
    }
  }
  if (vw.size() < sz) {
    if (p1 < v.size()) {
      for (int i = vw.size(); i < sz; i++) {
        vw.push_back(v[p1]);
        p1++;
      }
    } else {
      for (int i = vw.size(); i < sz; i++) {
        vw.push_back(w[p2]);
        p2++;
      }
    }
  }
}
void DFS(int v) {
  mark[v] = true;
  for (int i = 0; i < g[v].size(); i++) {
    if (mark[g[v][i]] == 0) {
      dis[g[v][i]] = dis[v] + 1;
      dp[g[v][i]][0] = v;
      for (int j = 0; j < pdd[v].size() && j < 10; j++)
        pd[g[v][i]][0].push_back(pdd[v][j]);
      for (int j = 1; j <= lgn; j++) {
        dp[g[v][i]][j] = dp[dp[g[v][i]][j - 1]][j - 1];
        merge(pd[g[v][i]][j - 1], pd[dp[g[v][i]][j - 1]][j - 1],
              pd[g[v][i]][j]);
      }
      DFS(g[v][i]);
    }
  }
}
int par(int v, int p) {
  while (p > 0) {
    int pp = p & (-p);
    v = dp[v][lg(pp)];
    p -= pp;
  }
  return v;
}
vector<int> par2(int v, int p) {
  vector<int> ans;
  while (p > 0) {
    int pp = p & (-p);
    vector<int> anss;
    merge(ans, pd[v][lg(pp)], anss);
    while (ans.size()) ans.pop_back();
    for (int i = 0; i < anss.size(); i++) ans.push_back(anss[i]);
    v = dp[v][lg(pp)];
    p -= pp;
  }
  return ans;
}
int LCA(int u, int v, int x) {
  if (u == v) return u;
  for (int i = x; i >= 0; i--) {
    if (dp[v][i] != dp[u][i]) return LCA(dp[v][i], dp[u][i], i - 1);
  }
  return dp[u][0];
}
int main() {
  cin >> n >> m >> q;
  lgn = lg(n);
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[v - 1].push_back(u - 1);
    g[u - 1].push_back(v - 1);
  }
  for (int i = 1; i <= m; i++) {
    int c;
    cin >> c;
    pdd[c - 1].push_back(i);
  }
  DFS(0);
  while (q--) {
    int u, v, a;
    cin >> v >> u >> a;
    u--;
    v--;
    int uu = u, vv = v;
    if (dis[v] > dis[u]) vv = par(v, dis[v] - dis[u]);
    if (dis[v] < dis[u]) uu = par(u, dis[u] - dis[v]);
    int lca = LCA(uu, vv, lgn);
    vector<int> jav;
    merge(par2(v, dis[v] - dis[lca] - 1), par2(u, dis[u] - dis[lca]), jav);
    if (dis[v] - dis[lca] - 1 >= 0) merge(pdd[v], jav, jav);
    merge(pdd[u], jav, jav);
    int aa = jav.size();
    aa = min(a, aa);
    cout << aa << ' ';
    for (int i = 0; i < aa; i++) cout << jav[i] << ' ';
    cout << '\n';
  }
}